How do I bind a member function template passing unspecified call wrapper - c++

I tried to compile the following example using VC11 and g++ 4.7.2:
#include <functional>
class X {
public:
template <typename T>
explicit X(T t)
{
std::bind(&X::invoke<T>, this, t)();
}
private:
template <typename T>
void invoke(T t)
{
t();
}
};
class Y {
public:
void foo() {
//...
}
};
int main() {
Y y;
X x(std::bind(&Y::foo, &y));
return 0;
}
but it finished with errors. I'm not sure if it is reasonable to paste the whole compilers output but generally
vc11 says:
error C2664: 'void std::_Pmf_wrap::operator ()(_Farg0 &,_V0_t) const' : cannot convert parameter 3 from 'void' to 'std::_Bind,Y *,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil,std::_Nil>' c:\program files (x86)\microsoft visual studio 11.0\vc\include\functional 1152 1 ConsoleApplication1 (Microsoft Visual C++ Compiler Nov 2012 CTP)
and g++:
Compilation finished with errors:
source.cpp: In instantiation of 'X::X(T) [with T = std::_Bind(Y*)>]':
source.cpp:28:33: required from here
source.cpp:8:9: error: no match for call to '(std::_Bind_helper(Y*)>), X* const, std::_Bind(Y*)>&>::type {aka std::_Bind(Y*)>)>(X*, std::_Bind(Y*)>)>}) ()'
Is there a way to solve this issue. It's very important for me, to save the main idea - a class that can be instantiated with any callable object (function object, function pointer or call wrapper returned by std::bind() function).
I would be grateful if someone help.
P.S. It compiles if I create an instance of X, passing function object or function pointer.

I think they omitted an important bit of boost::bind when adopting it into std::bind, namely boost::protect(). Your code can be fixed like the following:
#include <boost/bind/protect.hpp>
// ...
X x(boost::protect(std::bind(&Y::foo, &y)));
Or, alternatively:
template <typename T>
explicit X(T t)
{
auto tt = boost::protect(t);
auto f = std::bind(&X::invoke<decltype(tt)>, this, tt);
f();
}
See http://www.boost.org/doc/libs/1_53_0/libs/bind/bind.html
Although the first argument is, by default, not evaluated, all other arguments are. Sometimes it is necessary not to evaluate arguments subsequent to the first, even when they are nested bind subexpressions. This can be achieved with the help of another function object, protect, that masks the type so that bind does not recognize and evaluate it. When called, protect simply forwards the argument list to the other function object unmodified.
The header boost/bind/protect.hpp contains an implementation of protect. To protect a bind function object from evaluation, use protect(bind(f, ...)).
The upcoming Effective C++11: Content and Status by Scott Meyers is going to recommend prefer lambdas to std::bind. In C++11 you can simply do:
template <typename T>
explicit X(T t)
{
auto f = [t, this]() { this->invoke(t); };
f();
}
// ...
X x([&y](){ y.foo(); });

The root cause of the problem seems to be the internal copying of its arguments performed by std::bind, with particular reference to t.
You may want to workaround it this way:
template <typename T>
explicit X(T t)
{
std::bind(&X::invoke<T>, this, std::placeholders::_1)(t);
// ^^^^^^^^^^^^^^^^^^^^^ ^
}
This will also work, but you won't be allowed to make the result of bind outlive the argument t (otherwise, you would be passing a dangling reference to invoke<T>()):
template <typename T>
explicit X(T t)
{
std::bind(&X::invoke<T>, this, cref(t))();
// ^^^^^^^
}
UPDATE:
The above solutions are workarounds that help for achieving what you are showing in your example. However, it emerged from the comments that your use case may be quite different (e.g. passing around the result of bind for later evaluation).
As correctly pointed out by Maxim Yegorushkin in his answer, the conceptually correct solution consists in using a construct such as Boost's protect.
In case you do not want to use Boost, it is quite easy to define your own protect() function using C++11's variadic templates:
// Imitates boost::protect, but with variadic templates and perfect forwarding
namespace detail
{
template<typename F>
struct protect
{
private:
F _f;
public:
explicit protect(F f): _f(f)
{
}
template<typename... Ts>
auto operator () (Ts&&... args) ->
decltype(_f(std::forward<Ts>(args)...))
{
return _f(std::forward<Ts>(args)...);
}
};
}
template<typename F>
detail::protect<F> protect(F&& f)
{
return detail::protect<F>(std::forward<F>(f));
}
Eventually, this is how you could use it in your class, as Maxim suggested:
class X
{
public:
template <typename T>
explicit X(T t)
{
auto pt = protect(t);
std::bind(&X::invoke<decltype(pt)>, this, pt)();
}
private:
template <typename T>
void invoke(T t)
{
t();
}
};

Related

Passing a std::shared_ptr<T> to a function that takes a std::shared_ptr<const T>?

I have a function that needs to take shared ownership of an argument, but does not modify it.
I have made the argument a shared_ptr<const T> to clearly convey this intent.
template <typename T>
void func(std::shared_ptr<const T> ptr){}
I would like to call this function with a shared_ptr to a non-const T. For example:
auto nonConstInt = std::make_shared<int>();
func(nonConstInt);
However this generates a compile error on VC 2017:
error C2672: 'func': no matching overloaded function found
error C2784: 'void func(std::shared_ptr<const _Ty>)': could not deduce template argument for 'std::shared_ptr<const _Ty>' from 'std::shared_ptr<int>'
note: see declaration of 'func'
Is there a way to make this work without:
Modifying the calls to func. This is part of a larger code refactoring, and I would prefer not to have to use std::const_pointer_cast at every call site.
Defining multiple overloads of func as that seems redundant.
We are currently compiling against the C++14 standard, with plans to move to c++17 soon, if that helps.
template <typename T>
void cfunc(std::shared_ptr<const T> ptr){
// implementation
}
template <typename T>
void func(std::shared_ptr<T> ptr){ return cfunc<T>(std::move(ptr)); }
template <typename T>
void func(std::shared_ptr<const T> ptr){ return cfunc<T>(std::move(ptr)); }
this matches how cbegin works, and the "overloads" are trivial forwarders with nearly zero cost.
Unfortunately, there is no good solution to what you desire. The error occurs because it fails to deduce template argument T. During argument deduction it attempts only a few simple conversations and you cannot influence it in any way.
Think of it: to cast from std::shared_ptr<T> to some std::shared_ptr<const U> it requires to know U, so how should compiler be able to tell that U=T and not some other type? You can always cast to std::shared_ptr<const void>, so why not U=void? So such searches aren't performed at all as in general it is not solvable. Perhaps, hypothetically one could propose a feature where certain user-explicitly-declared casts are attempted for argument deduction but it isn't a part of C++.
Only advise is to write function declaration without const:
template <typename T>
void func(std::shared_ptr<T> ptr){}
You could try to show your intent by making the function into a redirection like:
template <typename T>
void func(std::shared_ptr<T> ptr)
{
func_impl<T>(std::move(ptr));
}
Where func_impl is the implementation function that accepts a std::shared_ptr<const T>. Or even perform const cast directly upon calling func_impl.
Thanks for the replies.
I ended up solving this a slightly different way. I changed the function parameter to just a shared_ptr to any T so that it would allow const types, then I used std::enable_if to restrict the template to types that I care about. (In my case vector<T> and const vector<T>)
The call sites don't need to be modified. The function will compile when called with both shared_ptr<const T> and shared_ptr<T> without needing separate overloads.
Here's a complete example that compiles on VC, GCC, and clang:
#include <iostream>
#include <memory>
#include <vector>
template<typename T>
struct is_vector : public std::false_type{};
template<typename T>
struct is_vector<std::vector<T>> : public std::true_type{};
template<typename T>
struct is_vector<const std::vector<T>> : public std::true_type{};
template <typename ArrayType,
typename std::enable_if_t<is_vector<ArrayType>::value>* = nullptr>
void func( std::shared_ptr<ArrayType> ptr) {
}
int main()
{
std::shared_ptr< const std::vector<int> > constPtr;
std::shared_ptr< std::vector<int> > nonConstPtr;
func(constPtr);
func(nonConstPtr);
}
The only downside is that the non-const instantiation of func will allow non-const methods to be called on the passed-in ptr. In my case a compile error will still be generated since there are some calls to the const version of func and both versions come from the same template.
As the const is only for documentation, make it a comment:
template <class T>
void func(std::shared_ptr</*const*/ T> p) {
}
You could additionally delegate to the version getting a pointer to constant object if the function is hefty enough to make it worthwhile:
template <class T>
void func(std::shared_ptr</*const*/ T> p) {
if (!std::is_const<T>::value) // TODO: constexpr
return func<const T>(std::move(p));
}
No guarantee the compiler will eliminate the move though.
You certainly don't want to be modifying the call sites, but you sure can be modifying the functions themselves - that's what you implied in the question anyway. Something had to be changed somewhere, after all.
Thus:
In C++17 you could use deduction guides and modify call sites, but in a less intrusive manner than with a cast. I'm sure a language lawyer can pitch in about whether adding a deduction guide to the std namespace is allowed. At the very least we can limit the applicability of those deduction guides to the types we care about - it won't work for the others. That's to limit potential for mayhem.
template <typename T>
class allow_const_shared_ptr_cast : public std::integral_constant<bool, false> {};
template <typename T>
static constexpr bool allow_const_shared_ptr_cast_v = allow_const_shared_ptr_cast<T>::value;
template<>
class allow_const_shared_ptr_cast<int> : public std::integral_constant<bool, true> {};
template <typename T>
void func(std::shared_ptr<const T> ptr) {}
namespace std {
template<class Y> shared_ptr(const shared_ptr<Y>&) noexcept
-> shared_ptr<std::enable_if_t<allow_const_shared_ptr_cast_v<Y>, const Y>>;
template<class Y> shared_ptr(shared_ptr<Y>&&) noexcept
-> shared_ptr<std::enable_if_t<allow_const_shared_ptr_cast_v<Y>, const Y>>;
}
void test() {
std::shared_ptr<int> nonConstInt;
func(std::shared_ptr(nonConstInt));
func(std::shared_ptr(std::make_shared<int>()));
}
std::shared_ptr is certainly less wordy than std::const_pointer_cast<SomeType>.
This should not have any performance impact, but sure modifying the call sites is a pain.
Otherwise there's no solution that doesn't involve modifying the called function declarations - but the modification should be acceptable, I think, since it's not any more wordy than what you had already:

Compiler infering the template argument

template<typename T>
class A
{
public:
A(T &t)
: t_(t){}
T t_;
};
int main()
{
int value;
A<decltype(value)> a(value);
// what I wish for : A a(value);
// which does not compile "missing template argument before 'a'"
}
Is there a way in the declaration of A (or somewhere else) to hint the compiler that T should be automatically resolved to the type passed to the constructor ?
(ideally c++11, but happy to hear about less old versions)
C++17 does it out of the box (or with the help of deduction guides), previous versions cannot.
As answered by #Quentin, this is only possible starting in C++17. However, if you're fine with calling a function to create your A objects, the following should do what you want in C++11:
template <class T, class NonRefT = typename std::remove_reference<T>::type>
A<NonRefT> create_A (T && t) {
return A<NonRefT>(std::forward<T>(t));
}
// Or even closer to your original code:
template <class T>
auto create_A (T && t) -> A<decltype(t)> {
return A<decltype(t)>(std::forward<T>(t));
}
I used std::remove_reference based on your use of decltype, though you might want to use std::decay instead.
int main () {
int value = 5;
auto a = create_A(value);
}
If I remember correctly the example code has an edge-case where it does not work as expected pre-C++17. The compiler will elide the copy/move constructor to create a from the rvalue returned by create_A(). However, it will check during compilation whether A's copy/move constructor (which it won't use) is available/accessible. Starting from C++17 the copy/move elision is done "properly" and no copy/move constructor is necessary for such code. (Also, I might be misremembering and it might be checking for copy/move assignment instead.)
In C++11 you can create a simple make_A function like this:
#include <iostream>
template <typename T>
class A {
public:
A(T &t) : t_(t) {}
T t_;
};
template <typename T>
A<T> make_A(T&& t) {
return A<T>(std::forward<T>(t));
}
int main() {
int value = 0;
auto a = make_A(value);
return 0;
}
Demo

Template parameter can't be deduced on implicitly constructed argument

I would like to have the following code in c++17:
#include <iostream>
#include <string>
#include <type_traits>
#include <functional>
class Foo;
template<class T>
class Bar {
public:
std::function<T(Foo&)> m_fn;
template<class Fn>
Bar(Fn fn) : m_fn(fn) {};
T thing(Foo &foo) const {
return m_fn(foo);
}
};
template<class Fn>
Bar(Fn) -> Bar<decltype(std::invoke(std::declval<Fn>(),
std::declval<Foo&>()))>;
class Foo {
public:
Foo() {};
template<class T>
std::vector<T> do_thing(const Bar<T> &b) {
std::vector<T> r;
r.push_back(b.thing(*this));
return r;
}
};
std::string test(Foo &) {
return "hello";
}
int main() {
Foo foo = Foo();
// works
std::vector<std::string> s = foo.do_thing(Bar{test});
// cant deduce T parameter to do_thing
std::vector<std::string> s = foo.do_thing({test});
}
But compiling this gives me "couldn't deduce template parameter ‘T’" on the call to do_thing.
Having do_thing(Bar{test}) fixes this and works fine but equates to some ugly code in the real code equivalent. I would like to have do_thing({test}) or do_thing(test) implicitly construct a Bar and pass that as the argument if possible.
I also don't want to forward declare a variable to pass into do_thing either
Is there some way to guide the inference of template argument T so that the call to do_thing can stay clean?
Edit:
Sorry for the late edit, but the arguments to the Bar constructor are over simplified in the example I included. In reality, there is an extra parameter std::optional<std::string> desc = std::nullopt and that might change in the future (although unlikely). So constructing the Bar inside do_thing would be a bit hard to maintain...
would like to have do_thing({test}) or do_thing(test) implicitly construct a Bar and pass that as the argument if possible.
Unfortunately, when you call do_thing({test}) or do_thing(test), test (or {test}) isn't a Bar<T> object. So the compiler can't deduce the T type and can't construct a Bar<T> object.
A sort of chicken-and-egg problem.
The best I can imagine is to add, in Foo, a do_test() method as follows
template<typename T>
auto do_thing (T const & t)
{ return do_thing(Bar{t}); }
This way you can call (without graphs)
std::vector<std::string> s = foo.do_thing(test);
You get the same result as
std::vector<std::string> s = foo.do_thing(Bar{test});
-- EDIT --
The OP ask
is there any way of preserving the {test} brace syntax? maybe with initializer_list or something?
Yes... with std::initializer_list
template<typename T>
auto do_thing (std::initializer_list<T> const & l)
{ return do_thing(Bar{*(l.begin())}); }
but, this way, you accept also
std::vector<std::string> s = foo.do_thing(Bar{test1, test2, test3});
using only test1
Maybe a little better... another way can be through a C-style array
template <typename T>
auto do_thing (T const (&arr)[1])
{ return do_thing(arr[0]); }
This way you accept only an element.
This happens because {} is not an expression and can only be used in limited ways while doing argument deduction, the parameter must have specific forms in order to succeed.
The allowed parameters types that can be used to deduce template parameters when {} is involved are better expanded in [temp.deduct.call]/1, two of the examples extracted from the cited part of the standard are:
template<class T> void f(std::initializer_list<T>);
f({1,2,3}); // T deduced to int
template<class T, int N> void h(T const(&)[N]);
h({1,2,3}); // T deduced to int
In your example the deduction guide is not used to deduce the T for {test} for the same as above.
foo.do_thing(Bar{test});
is your direct option without using additional functions.

How can I get the C++ compiler to deduce T indirectly?

My template-fu is rather weak. I have this code:
template<typename T>
void Foo(void(*func)(T*)) { }
void Callback(int* data) { }
int Test()
{
Foo(Callback);
}
...but I'd like something more readable than C's nasty function pointer syntax of void(*func)(T*).
Someone on my team suggested this:
template<typename T>
struct Types
{
typedef void Func(T*);
};
template<typename T>
void Foo2(typename Types<T>::Func* func) {}
void Test2()
{
Foo2(Callback); // could not deduce template argument for 'T'
Foo2<int>(Callback); // ok
}
(I'm still debating whether this is actually more readable, but that's a separate issue.)
How can I help the compiler figure out what T is without needing to explicitly specify it in the caller?
You can extract T from the function type using a traits class.
template<class F>
struct CallbackTraits;
template<class T>
struct CallbackTraits<void(*)(T)>
{
typedef T ArgumentType;
};
Your example can be modified like this:
template<typename F>
void Foo(F func)
{
typedef typename CallbackTraits<F>::ArgumentType T;
}
void Callback(int* data) { }
int Test()
{
Foo(Callback);
}
This technique is used in the boost type-traits library:
http://www.boost.org/doc/libs/1_57_0/libs/type_traits/doc/html/boost_typetraits/reference/function_traits.html
This blog post goes into a bit more detail about the implementation of the technique:
https://functionalcpp.wordpress.com/2013/08/05/function-traits/
Unfortunately this approach hides the information in the signature of Foo about the constraints on the argument passed in. In the above example the argument must be a function of type void(T*).
This alternative syntax does the same as the original example while being slightly more readable:
template<typename T>
void Foo(void func(T*)) { }
Another alternative syntax that may be more readable can be achieved using c++11's alias templates as follows:
template<typename T>
using Identity = T;
template<typename T>
void Foo(Identity<void(T*)> func) { }
Unforunately the latest MSVC fails to compile this, reporting an internal compiler error.
You won't be able to deduce the type based on a nested name: there is no reason why different instantiations of the outer type won't define an identical inner type. You could use a using alias, though:
template <typename T>
using Function = auto (*)(T*) -> void;
template <typename T>
void Foo(Function<T>) {
}
Personally, I would recommend against using any of that, however: in practice it seems much more advisable to actually take a function object which later allows using object with suitable function call operators to be used. For callbacks it is quite common that you'll need to pass in some auxiliary data. That is, you would either use an unconstrained template or one which takes a type-erased type, depending on what you want to do exactly:
template <typename Fun>
void Unconstrained(Fun fun) {
}
template <typename T>
void TypeErased(std::function<void(T*)> fun) {
}
The unconstrained version has the advantage that it can potentially inline the function call but it has the disadvantage that every function object type creates a new instantiation and that the argument types are likely to vary. The type-erased version effectively has to do something like a virtual function call but there is just one instantiation of the function template (per argument type T, of course).
Admittedly, the type-erased version's type won't be deduced from a function pointer (or any other argument which isn't a std::function<void(X*)>), i.e., you may want to have a forwarding function
template <typename T>
void TypeErased(Function<T> fun) {
TypeErased(std::function<void(T)>(fun));
}
In C++98 and C++03 template argument deduction only works with functions (and methods).
I don't think the picture changed in the more recent standards.

Explicit specialization using variadic parameters within variadic templated class [MSVS '12: Nov. '12 CTP: error C3522]

I'm trying to expand a class's variadic template type list within a child method as such:
template<typename... P>
struct Foo
{
template<P...> // error C3522: 'P' : parameter
// pack cannot be expanded in this context
static void Bar(P... a){}
};
What is wrong with this code, or is it just a MSVS '12: Nov. '12 CTP bug?
(Yes, I know the explicit template specialization in this example is redundant.)
The above is the simplest case that I get to reproduce the error. The full code is:
template<typename FuncSignature>
class Callback;
template<typename R, typename... P>
class Callback<R (P...)>
{
public:
Callback() : func(0), obj(0) {}
Callback& operator=(const Callback& rhs)
{ obj = rhs.obj; func = rhs.func; return *this; }
private:
typedef R (*FuncType)(const void*, P...);
Callback(FuncType f, const void* o) : func(f), obj(o) {}
private:
FuncType func;
const void* obj;
template<typename FR, typename... FP>
friend class FreeCallbackFactory;
};
template<typename R, typename... P>
class FreeCallbackFactory
{
private:
template<R (*Func)(P...)>
static R Wrapper(const void*, P... a)
{
return (*Func)(a...);
}
public:
template<R (*Func)(P...)>
inline static Callback<R (P...)> Bind()
{
return Callback<R (P...)>
(&FreeCallbackFactory::Wrapper<Func>, 0);
}
};
template<typename R, typename... P>
inline FreeCallbackFactory<R, P...>
GetCallbackFactory(R (*)(P...))
{
return FreeCallbackFactory<R, P...>();
}
void Test(){}
int main(int argc, char** argv){
Callback<void ()> cb = GetCallbackFactory(&Test).Bind<&Test>()
}
It compiles fine in g++ so I'm assuming just a compiler bug, and continued findings still only point to this, are there any possible workarounds for this other than explicitly expanding them out one by one?
Edit: This has been reported to the compiler team as a bug and a patch will be in the next release of the compiler. [Link]
The code looks correct put I doubt that it does what you intended it to do: The declaration
template <P...>
static void Bar(P... a);
declares a function taking P... values as template argument and as function argument. That is, the elements of P need to be of a type allowing non-type parameters (e.g., integers, pointers, or references) and you'd need to provide their respective values as template parameters when calling the function. The call to a function like this would look something like this (although it seems neither gcc nor clang require the template parameters to be passed):
Foo<int, int>::Bar<1, 2>(3, 4);
That said, based on the error messages generated by both gcc and clang it seems that they won't let you create a specialization of member function templates but I haven't verified this in the standard. I think, you should probably just leave out the template declaration and use
static void Bar(P... a);
To me the code does not look correct. The pack expansion P... expands to a set of types. Depending on what you aim to achieve the correct thing to do would be to remove the template<P...> part, in which case each class template instance has one Bar() method with the same parameters as the class. The other would be that this is an unrelated parameter pack in which case an unbounded set of methods is generated, in that case you must declare the method something like:
template<typename... U> static void Bar(U...);
reusing the name P would not be very useful, I guess.